Explanation:

To solve this problem, we can use the principle of virtual work. The horizontal thrust developed by the supports can be determined by considering the equilibrium of virtual forces.

Equilibrium of virtual forces:

Let's consider a small element of the cable between points A and B. This element has a length dx and is located at a distance x from point A.

The weight of this element is given by dW = w*dx, where w is the uniformly distributed load per unit length.

The tension in the cable at point A is T_A and at point B is T_B.

The horizontal thrust developed by the supports can be determined by considering the equilibrium of virtual forces. We can assume a virtual displacement of the cable, denoted by dy, in the horizontal direction.

The virtual work done by the weight of the element is given by dW * dy = w*dx*dy.

The virtual work done by the tension at point A is T_A * dy.

The virtual work done by the tension at point B is T_B * dy.

Since the cable is in equilibrium, the sum of the virtual work done by all these forces should be zero.

Equilibrium equation:

Summing up the virtual work done by all forces, we have:

w*dx*dy + T_A * dy + T_B * dy = 0

Substituting the expressions for T_A and T_B, we have:

w*dx*dy + T_A * dy + T_A * (l - x) * dy = 0

Simplifying the equation:

(w*dx + 2*T_A*dy) * dy = 0

Since this equation should hold for any arbitrary values of dx and dy, the coefficient of dy should be zero.

Therefore, we have the following equation:

w*dx + 2*T_A*dy = 0

Horizontal thrust equation:

Solving this equation for T_A, we get:

T_A = -w*dx / (2*dy)

The horizontal thrust developed by the supports is equal to the tension at point A, T_A. Therefore, the horizontal thrust is given by:

Horizontal thrust = -w*dx / (2*dy)

To find the total horizontal thrust, we need to integrate this expression over the entire length of the cable, from x = 0 to x = l.

Integrating both sides of the equation, we get:

Total horizontal thrust = -w * integral(dx / (2*dy), x = 0 to x = l)

Simplifying the integral, we have:

Total horizontal thrust = -w * [ln(dy) / (2*dy)] (evaluated from x = 0 to x = l)

Since the cable is symmetrical and the central sag is h, we have dy = h + y, where y is the vertical displacement of the cable at distance x from point A.

Substituting this expression for dy, we have:

Total horizontal thrust = -w * [ln(h + y) / (2*(h + y))] (evaluated from x = 0 to x = l)

To find the horizontal thrust at the supports, we need to evaluate this expression at y = h.

Substituting y = h, we have:

Total horizontal thrust = -w * [ln(h + h) / (2*(h + h))] (evaluated from x =